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Next: Exercises Up: Magnetostatic Fields Previous: Circular Current Loop

Localized Current Distribution

Consider the magnetic field generated by a current distribution that is localized in some relatively small region of space centered on the origin. From Equation (621), we have

$\displaystyle {\bf A}({\bf r}) = \frac{\mu_0}{4\pi}\int\frac{{\bf j}({\bf r}')}{\vert{\bf r}-{\bf r}'\vert}\,dV'.$ (661)

Assuming that $ r\gg r'$ , so that our observation point lies well outside the distribution, we can write

$\displaystyle \frac{1}{\vert{\bf r}-{\bf r}'\vert}= \frac{1}{\vert{\bf r}\vert}+\frac{{\bf r}\cdot{\bf r}'}{\vert{\bf r}\vert^{\,3}}+\cdots.$ (662)

Thus, the $ i$ th Cartesian component of the vector potential has the expansion

$\displaystyle A_i({\bf r})= \frac{\mu_0}{4\pi}\,\frac{1}{\vert{\bf r}\vert}\int...
...{{\bf r}}{\vert{\bf r}\vert^{\,3}}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV'+\cdots$ (663)

Consider the integral

$\displaystyle K = \int (f\,{\bf j}\cdot \nabla' g+g\,{\bf j}\cdot\nabla' f)\,dV',$ (664)

where $ {\bf j}({\bf r}')$ is a divergence-free [see Equation (618)] localized current distribution, and $ f({\bf r}')$ and $ g({\bf r}')$ are two well-behaved functions. Integrating the first term by parts, making use of the fact that $ {\bf j}'({\bf r}')\rightarrow {\bf0}$ as $ \vert{\bf r}'\vert\rightarrow\infty$ (because the current distribution is localized), we obtain

$\displaystyle K=\int\left[-g\,\nabla'\cdot(f\,{\bf j})+g\,{\bf j}\cdot\nabla' f\right]dV'$ (665)


$\displaystyle K = \int\left[-g\,{\bf j}\cdot\nabla' f -g\,f\,\nabla'\cdot{\bf j}+g\,{\bf j}\cdot\nabla' f\right]dV'=0,$ (666)

because $ \nabla'\cdot{\bf j}= 0$ . Thus, we have proved that

$\displaystyle \int (f\,{\bf j}\cdot \nabla' g+g\,{\bf j}\cdot\nabla' f)\,dV'=0.$ (667)

Let $ f=1$ and $ g=x_i'$ (where $ x_i'$ is the $ i$ th component of $ {\bf r}'$ ). It immediately follows from Equation (668) that

$\displaystyle \int j_i({\bf r}')\,dV'= 0.$ (668)

Likewise, if $ f=x_i'$ and $ g=x_j'$ then Equation (668) implies that

$\displaystyle \int \left(x_i'\,j_j + x_j'\,j_i\right)\,dV' = 0.$ (669)

According to Equations (664) and (669),

$\displaystyle A_i({\bf r})= \frac{\mu_0}{4\pi}\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV'+\cdots.$ (670)


$\displaystyle {\bf r}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV' = x_j\int x'_j\,j_i\,dV'= -\frac{1}{2}x_j\int(x_i'\,j_j-x_j'\,j_i)\,dV',$ (671)

where use has been made of Equation (670), as well as the Einstein summation convention. Thus,

$\displaystyle {\bf r}\cdot\int j_i\,{\bf r}'\,dV' = -\frac{1}{2}\int \left[({\b...
... = -\frac{1}{2}\left[{\bf r}\times \int ({\bf r}'\times {\bf j})\,dV'\right]_i.$ (672)

Hence, we obtain

$\displaystyle {\bf A}({\bf r})=-\frac{\mu_0}{8\pi}\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}}\times \int ({\bf r}'\times {\bf j})\,dV'+\cdots.$ (673)

It is conventional to define the magnetization, or magnetic moment density, as

$\displaystyle {\bf M}({\bf r}) = \frac{1}{2}\,{\bf r}\times {\bf j}({\bf r}).$ (674)

The integral of this quantity is known as the magnetic moment:

$\displaystyle {\bf m} = \frac{1}{2}\int{\bf r}'\times {\bf j}'({\bf r}')\,dV'.$ (675)

It immediately follows from Equation (674) that the vector potential a long way from a localized current distribution takes the form

$\displaystyle {\bf A}({\bf r}) = \frac{\mu_0}{4\pi}\,\frac{{\bf m}\times {\bf r}}{r^{\,3}}.$ (676)

The corresponding magnetic field is

$\displaystyle {\bf B}({\bf r}) = \nabla\times {\bf A} = \frac{\mu_0}{4\pi}\left[\frac{3\,({\bf m}\cdot{\bf r})\,{\bf r} -r^{\,2}\,{\bf m}}{r^{\,5}}\right].$ (677)

Thus, we have demonstrated that the magnetic field far from any localized current distribution takes the form of a magnetic dipole field whose moment is given by the integral (676).

Consider a localized current distribution that consists of a closed planar loop carrying the current $ I$ . If $ d{\bf r}$ is a line element of the loop then Equation (676) reduces to

$\displaystyle {\bf m} = I\oint\frac{1}{2}\,{\bf r}\times d{\bf r}.$ (678)

However, $ (1/2)\,{\bf r}\times d{\bf r}=d{\bf A}$ , where $ d{\bf A}$ is a triangular element of vector area defined by the two ends of $ d{\bf r}$ and the origin. Thus, the loop integral gives the total vector area, $ {\bf A}$ , of the loop. It follows that

$\displaystyle {\bf m} = I\,A\,{\bf n},$ (679)

where $ {\bf n}$ is a unit normal to the loop in the sense determined by the right-hand circulation rule (with the current determining the sense of circulation). Of course, Equation (680) is identical to Equation (659).

next up previous
Next: Exercises Up: Magnetostatic Fields Previous: Circular Current Loop
Richard Fitzpatrick 2014-06-27